What are the solutions to the equation 25x^4+4=29x^2 ?

The solutions to the equation 25x^4+4=29x^2 are x=1,x=-1,x= 2/5 ,x=-2/5

Find the zeros of 25x^4+4=29x^2

The zeros of 25x^4+4=29x^2 are x=1,x=-1,x= 2/5 ,x=-2/5

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25 x^{4} + 4 = 29 x^{2}

x = 1, x = - 1, x = \frac{2}{5}, x = - \frac{2}{5}

Decimal Notation

x = 1, x = - 1, x = 0.4, x = - 0.4

Solution steps

25 x^{4} + 4 = 29 x^{2}

Move

29 x^{2}

to the left side

25 x^{4} + 4 - 29 x^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

25 x^{4} - 29 x^{2} + 4 = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

25 u^{2} - 29 u + 4 = 0

Solve

25 u^{2} - 29 u + 4 = 0 : u = 1, u = \frac{4}{25}

u = 1, u = \frac{4}{25}

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = 1 : x = 1, x = - 1

Solve

x^{2} = \frac{4}{25} : x = \frac{2}{5}, x = - \frac{2}{5}

The solutions are

x = 1, x = - 1, x = \frac{2}{5}, x = - \frac{2}{5}

8 x^{3} + 3 x^{2} - 16 x - 1 = 0

x = x^{5}

n^{3} - n^{2} - 16 n - 20 = 0

x^{3} - 8 x^{2} - 15 = 0

(x - 1) \cdot x \cdot (x + 1) = 210

What are the solutions to the equation 25x^4+4=29x^2 ?
The solutions to the equation 25x^4+4=29x^2 are x=1,x=-1,x= 2/5 ,x=-2/5
Find the zeros of 25x^4+4=29x^2
The zeros of 25x^4+4=29x^2 are x=1,x=-1,x= 2/5 ,x=-2/5